Example. Why the periodic trace needs gauge projection [fdip-0084]

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Take a graph with one edge and two distinct vertices, with \(G=\mathbb Z_2\), \(V=0\) and the sign kernel \(k(u)=e^{\beta u}\), \(\beta >0\). The link space has orthonormal basis \(1,u\), with \[T1=(\cosh \beta )1,\qquad Tu=(\sinh \beta )u.\] The two endpoint gauge transformations act by \(u\mapsto h_tuh_s\). Their action is transitive on the two link configurations. Thus the gauge projector satisfies \(P1=1\) and \(Pu=0\), and the physical space consists exactly of constants.

For every \(N\ge 1\), the projected and unprojected traces are \[\begin {aligned} \operatorname {Tr}(PT^N)&=(\cosh \beta )^N,\\ \operatorname {Tr}(T^N)&=(\cosh \beta )^N+(\sinh \beta )^N. \end {aligned}\] They are different because \(\beta >0\). In the temporal-link integral, each independent endpoint average gives \[\frac 14\sum _{h_s,h_t=\pm 1} e^{\beta U h_sU'h_t}=\cosh \beta .\] The remaining normalized link integrals contribute one, reproducing the projected answer. Dropping \(P\) would retain the nonphysical sign mode and change this finite periodic law.